# Question 2997a

May 28, 2016

$\text{2.5 g sucrose}$

#### Explanation:

Assuming that you're dealing with percent concentration by mass, $\text{% m/m}$, you can determine the mass of sucrose present in $\text{1.0 kg}$ of that solution by using its concentration as a conversion factor.

This is essentially what a solution's concentration does -- it helps you go from, in this case, mass of solution to mass of solute or vice versa.

A 0.25% by mass solution will contain $\text{0.25 g}$ of sucrose, your solute, for every $\text{100 g}$ of solution. So, if every $\text{100 g}$ of this solution will contain $\text{0.25 g}$ of sucrose, it follows that $\text{1.0 kg}$ will contain

1.0 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * overbrace("0.25 g sucrose"/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 0.25% m/m")) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.5 g sucrose")color(white)(a/a)|)))#